Add ammonia, hydroxide or concentrated hydrochloric acid to a solution of a transition metal ion and the ligands around the metal are replaced. The colour changes, and sometimes a precipitate appears and then redissolves, giving a fingerprint for each ion.
In ligand exchange one ligand in a complex is replaced by another. Pale blue [Cu(H2O)6]2+ reacts with excess aqueous ammonia to give deep blue [Cu(NH3)4(H2O)2]2+; NH3 and H2O are similar in size, so the coordination number stays 6. With concentrated HCl it gives yellow-green [CuCl4]2−: the large Cl− ions change the coordination number to 4 and the shape to tetrahedral. Pink [Co(H2O)6]2+ similarly gives blue [CoCl4]2−. Adding a little NaOH(aq) or NH3(aq) removes H+ from water ligands and precipitates the neutral hydroxide, e.g. blue Cu(OH)2(H2O)4; some precipitates redissolve in excess reagent.
Follow each arrow on the figure: write the equation, then say what happens to the colour, the coordination number and the shape.
| Reagent | Equation | Colour change | Coordination number, shape |
|---|---|---|---|
| excess NH3(aq) | [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O | pale blue → deep blue | 6 → 6, octahedral |
| conc. HCl | [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O | pale blue → yellow-green | 6 → 4, octahedral → tetrahedral |
| conc. HCl | [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O | pink → blue | 6 → 4, octahedral → tetrahedral |
| excess NH3(aq) | [Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O | pink → pale brown (darkens in air as Co2+ is oxidised to Co3+) | 6 → 6, octahedral |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is seen when excess aqueous ammonia is added to aqueous copper(II) sulfate?
What colour is [CoCl4]2−?
Which ion gives a red-brown precipitate with aqueous sodium hydroxide?
Which hydroxide precipitate dissolves in excess aqueous sodium hydroxide?
Add aqueous sodium hydroxide or aqueous ammonia dropwise and then in excess. The colour of the precipitate identifies the ion.
| Ion | NaOH(aq), a little | NaOH(aq), excess | NH3(aq), a little | NH3(aq), excess |
|---|---|---|---|---|
| Cu2+ | pale blue ppt | insoluble | pale blue ppt | dissolves: deep blue solution |
| Fe2+ | green ppt, turns brown at surface in air | insoluble | green ppt | insoluble |
| Fe3+ | red-brown (orange-brown) ppt | insoluble | red-brown ppt | insoluble |
| Co2+ | blue ppt, turns pink on standing | insoluble | blue ppt | dissolves: pale brown solution |
| Mn2+ | off-white ppt, turns brown in air | insoluble | off-white ppt | insoluble |
| Cr3+ | grey-green ppt | dissolves: dark green solution | grey-green ppt | insoluble |
A little ammonia acts as a base and gives the hydroxide precipitate; only excess ammonia acts as a ligand and gives the deep blue solution. Concentrated HCl changes the coordination number from 6 to 4 because Cl− is large. [CuCl4]2− has a 2− charge (+2 + 4 × −1).
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
How do the coordination number and shape change when concentrated HCl is added to aqueous copper(II) ions?
Why does a little aqueous ammonia give a precipitate with Cu2+(aq) but excess gives a solution?
A green precipitate forms with NaOH(aq) and slowly turns brown at the surface. Which ion was present?
Which equation shows the formation of the precipitate when NaOH(aq) is added to aqueous iron(III) ions?
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